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Exam CV0-003 topic 1 question 101 discussion

Actual exam question from CompTIA's CV0-003

Question #: 101
Topic #: 1

An organization will be deploying a web application in a public cloud with two web servers, two database servers, and a load balancer that is accessible over a single public IP. Taking into account the gateway for this subnet and the potential to add two more web servers, which of the following will meet the minimum IP requirement?

A. 192.168.1.0/26
B. 192.168.1.0/27
C. 192.168.1.0/28
D. 192.168.1.0/29

Show Suggested Answer

Suggested Answer: C 🗳️

by achow26 at Sept. 7, 2022, 10:03 p.m.

Comments

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sweetykaur

4 months, 1 week ago

Selected Answer: B

To meet the minimum IP requirement for the described deployment scenario, you need to account for a total of 7 IP addresses (2 web servers + 2 database servers + 2 additional web servers + 1 load balancer). Additionally, you need to include the gateway and network/broadcast addresses, which means you'll need at least 11 IP addresses in total. The best option that meets these requirements is: B. 192.168.1.0/27 A /27 subnet provides 32 IP addresses (30 usable addresses after accounting for the network and broadcast addresses), which is more than sufficient for the deployment scenario and future expansion.

upvoted 1 times

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Pisces225

9 months, 1 week ago

Selected Answer: C

Some of you guys are going to fail your Net+ or anything else network related. Here's the proper breakout. C is indeed correct, but some of these comments are just terrible and no one has laid it out, so here it is: 1 - gateway (this will be our public IP but will have internal as well as all gateways always do!) 2 - load balancer 3 - web server #1 4 - web server #2 5 - database server 6 - database server 7 - new webserver #1 8 - new web server #2 9 - broadcast IP Our /28 is going to provide a maximum of 16 IPs which will cover the 9 addresses needed for current and future expansion.

upvoted 4 times

Stonetales987

8 months ago

10 - Network Address - The first address of each subnet is also unusable and is used to identify the subnet itself and cannot be assigned to a host.

upvoted 2 times

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db93ae3

10 months, 1 week ago

The load balancer will have a public IP. Doesn't that exclude it from the private network, leading to /29 being sufficient for 6 hosts

upvoted 1 times

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Not_That_Guy

2 years, 2 months ago

Selected Answer: C

7 addresses are needed, /28 provides 8 addresses but only 6 of those are available after discounting the network and broadcast addresses. So /29 is the smallest subnet that meets requirements.

upvoted 2 times

Not_That_Guy

2 years, 2 months ago

Oops, I reversed those. /29 provides 8 addresses; /28 is the smallest subnet that meets requirements.

upvoted 2 times

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ryanzou

2 years, 3 months ago

28 is correct

upvoted 2 times

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achow26

2 years, 4 months ago

Total IP requirement is 7 including future growth so a /29 can provide 8 IPs. So why a /28?

upvoted 1 times

concepcionz

1 year, 9 months ago

192.168.1.0/29 192.168.1.8/29 192.168.1.16/29 192.168.1.24/29 and so on So the Network is 192.168.1.0 and the Broadcast 192.168.1.7, that leaves 6 usable address (1,2,3,4,5,6)

upvoted 3 times

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i_bird

2 years, 3 months ago

/29 provides 6 not 8.. correct ans: /28 = 14 Ips

upvoted 5 times

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